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How to prove that the determinant of the product of matrices is equal to product of their respective determinants ?.ie,

$$ |A.B|=|A|.|B| $$ where $A,B$ are square matrices of the same order.

I have no clue of where to start, I just know the following $$ |A|I=A(adj A)=(adj A)A $$ Note: I have checked a similar problem at Proving determinant product rule combinatorially which gives a combinatorial proof of the statement which is kinda cumbersome by the use of Combinatorics in a matrix problem. So I hope its worth asking for an alternative proof using only the properties of matrices.


marked as duplicate by Hans Lundmark, GNUSupporter 8964民主女神 地下教會, Ali Caglayan, José Carlos Santos, Parcly Taxel Feb 15 '18 at 0:11

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    $\begingroup$ Usually one proves that any matrix can be written as a product of elementary matrices and then one shows that determinants are multiplicative on elementary matrices. $\endgroup$ – Mathematician 42 Feb 14 '18 at 11:05
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    $\begingroup$ If you want a rigorous proof of this, there are unfortunately no real shortcuts you can take. You can find a proof here with all the details: linear.ups.edu/html/section-PDM.html $\endgroup$ – 0XLR Feb 14 '18 at 11:08
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    $\begingroup$ Another approach is to define a function $f:\mathbb{R}^{n\times n}\to\mathbb{R}:X\to \det(AX)/\det(A)$ and show that this function has the property of being multilinear, alternating and maps the identity matrix to 1. It can be shown that the determinant function is the only map with these properties, thus $\det(X)=\det(AX)/\det(A)$. $\endgroup$ – Raskolnikov Feb 14 '18 at 11:11
  • $\begingroup$ @HansLundmark thanx for mentioning it. But i am looking for proofs only using properties of matrices. Hope its worth asking considering the combinatorial proof you mentioned is not an easy one. $\endgroup$ – ss1729 Feb 14 '18 at 12:41
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    $\begingroup$ It's o.k. not proving this in class. But deferring the proof (which is in all textbooks) to the exercises is cheating. $\endgroup$ – Christian Blatter Feb 14 '18 at 13:05

Let $A,B\in \Re^{n\times n}$ be two $n\times n$ matrices.

Now we can distinguish between 2 cases:

Case 1: $A$ isn't invertible. Therefore the product $AB$ won't be invertible either. In this case, $|AB|=|A|\cdot|B|=0$ holds true (both sides are equal to $0$).

Case 2: $A$ is invertible. In this case, both $A$ and the Identity Matrix $I_n$ are equivalent by rows: $$A=E_p\cdot E_{p-1}\ldots E_1\cdot I_n=E_p\cdot E_{p-1}\ldots E_1$$

And therefore: $$\begin{align*} |AB| &= |E_p\cdot E_{p-1}\ldots E_1\cdot B| \\ &=|E_p|\cdot|E_{p-1}\ldots E_1\cdot B| \\ &= |E_p|\cdot|E_{p-1}|\cdot |\ldots E_1\cdot B| \\ &=\ldots \\ &= |E_p\ldots E_1|\cdot |B| \\ &= |A|\cdot|B|\end{align*}$$

  • $\begingroup$ How do you show that $A$ singular implies $|A|=0$? $\endgroup$ – David C. Ullrich Feb 14 '18 at 15:14
  • $\begingroup$ That's by definition, if I'm not wrong. $A$ is singular iif $|A|=0$ and vice versa $\endgroup$ – Jose Lopez Garcia Feb 14 '18 at 17:48
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    $\begingroup$ The usual defiition is "not invertible" - in any case, above you use the fact that if $A$ is not invertible then $|A|=0$. How do you prove that? $\endgroup$ – David C. Ullrich Feb 14 '18 at 19:00

Notations like $\sum_j$ or $\prod_k$ will always mean $\sum_{j=1}^n$ and $\prod_{k=1}^n$; there's going to be enough typing here even with that convention.

If $\sigma\in S_n$ define $|\sigma|=1$ if $\sigma$ is even, $-1$ if $\sigma$ is odd. The simplest definition of the determinant is $$|A|=\sum_{\sigma\in S_n}|\sigma|\prod_ja_{j,\sigma(j)}.$$

Ok, maybe that doesn't look as simple as various other definitions, but it's often the simplest to use in proving things about determinants; if whatever formula is true it must follow by just plugging in and working it out. We're going to use this to show the determinant is multiplicative - the proof is "simple" in that it uses absolutely no previous results, nothing but the fact that multiplication distributes over addition.

In general what is the product of a bunch of sums, $\prod_j\sum_k\alpha_{jk}$? It's the sum of products, each product being the product of one term from the first sum times one term from the second sum, etc. That is, if we let $X$ be the set of all functions from $\{1,2,\dots,n\}$ to itself, in general we have $$\prod_j\sum_k\alpha_{jk} =\sum_{\psi\in X}\prod_j\alpha_{j,\psi(j)}.$$

We extend the notation $|\sigma|$ to $X$ by saying $|\sigma|=0$ if $\sigma\in X\setminus S_n$. Note that if $\sigma,\psi\in X$ and $\sigma\psi\in S_n$ then $\sigma$ must be surjective and $\psi$ must be injective, so in fact $\sigma,\psi\in S_n$. Hence $$|\sigma\psi|=|\sigma|\,|\psi|\quad(\sigma,\psi\in X).$$

Now say $AB=C$. In the obvious notation this says$$c_{jk}=\sum_ia_{ji}b_{ik}.$$So if $\sigma\in S_n$ then$$ \sum_jc_{j,\sigma(j)}=\sum_{\psi\in X}\prod_ja_{j,\psi(j)}b_{\psi(j),\sigma(j)}.$$

Hence $$|AB|=\sum_{\psi\in X}\sum_{\sigma\in S_n}|\sigma|\prod_ja_{j,\psi(j)}b_{\psi(j),\sigma(j)}.$$

Now note that since $|\psi|=0$ for $\psi\in X\setminus S_n$ we have $$|A|=\sum_{\psi\in X}|\psi|\prod_ja_{j,\psi(j)}.$$Multiplying this by the original definitoin of $|B|$ gives $$|A|\,|B|=\sum_{\psi\in X}\sum_{\phi\in S_n}|\sigma\psi|\prod_ja_{j,\psi((j)}b_{j,\sigma(j)}.$$

Since for every $\psi\in X$ either $|\psi\sigma|=0$ or $\psi$ is a permutation of $\{1,2\dots,n\}$ that last is equal to $$\sum_{\psi\in X}\sum_{\phi\in S_n}|\sigma\psi|\prod_ja_{j,\psi((j)}b_{\psi(j),\sigma(\psi(j))}.$$

And again, since if $|\psi|\ne0$ then $\sigma\mapsto\sigma\psi$ is a bijection on $S_n$, that last expression is equal to $$\sum_{\psi\in X}\sum_{\phi\in S_n}|\sigma|\prod_ja_{j,\psi((j)}b_{\psi(j),\sigma(j)},$$which is $|AB|$.

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    $\begingroup$ This is an unappropiate answer in my opinion. The OP said he has basic knowledge in combinatorics. He won't be able to understand a single line of math you wrote, with all due respect. Let's make maths simple and fun. $\endgroup$ – Jose Lopez Garcia Feb 14 '18 at 17:53
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    $\begingroup$ And I meant no disrespect by the way! I admire how much maths you know, for sure. But most of us, including OP, have limited knowledge and I'm sure this answer won't be any useful to him. You should seriously consider writing a book on advanced maths, though ;) $\endgroup$ – Jose Lopez Garcia Feb 14 '18 at 18:02
  • $\begingroup$ There's no law saying answers have to be solely for the benefit of the OP. This is a complete self-contained proof - the other answers so far all assume various preliminary not-quite-trivial things about determinants. $\endgroup$ – David C. Ullrich Feb 14 '18 at 19:10
  • $\begingroup$ @DavidC.Ullrich thnx. But a combinatorial proof is asked at math.stackexchange.com/questions/177560/…. I was looking for an alternative proof constrained to matrix algebra as much as possible. It'd be helpful if you could rather mention the drawbacks in other answer. $\endgroup$ – ss1729 Feb 14 '18 at 19:43
  • $\begingroup$ @ss1729 Sorry, I had no idea that the proof above was "combinatorial". $\endgroup$ – David C. Ullrich Feb 14 '18 at 19:51

Thanx @ZeroXLR for the hint.

$$ |A.B|=|A|.|B| $$

$\color{blue}{\text{Theorem 1}}:$ Suppose that $A$ is a square matrix of size $n$ and $E$ is any elementary matrix of size $n$, then $$\boxed{ |E.A|=|E|.|A|} $$

Proof :

1.) $A=[a_{i,j}]$ and $B=E_l(k).A=[b_{i,j}]$ where $b_{l,j}=k.a_{l,j}$

$$ |E_l(k).A|=|B|=\sum_{j=1}^{n} (-1)^{l+j}.b_{l,j}.|b(l\ |\ j)|\\ =\sum_{j=1}^{n} (-1)^{l+j}.b_{l,j}.|a(l\ |\ j)|\\ =\sum_{j=1}^{n} (-1)^{l+j}.k.a_{l,j}.|a(l\ |\ j)|\\ =k.\sum_{j=1}^{n} (-1)^{l+j}.a_{l,j}.|a(l\ |\ j)|\\ \color{red}{|E_l(k).A|=k.|A|=|E_l(k)|.|A|} $$ 2.) Swapping of adjascent rows, $A=[a_{i,j}]$ and $B=E_{l,l+1}.A=[b_{i,j}]$ where $a_{l,j}=b_{l+1,j}$ and $a_{l+1,j}=b_{l,j}$ $$ |E_{l,l+1}.A|=|B|=\sum_{j=1}^{n}(-1)^{l+1+j}.b_{l+1,j}.|b(l+1\ |\ j)|\\ =\sum_{j=1}^{n}(-1)^{l+j}.(-1)^1.a_{l,j}.|a(m\ |\ j)|\\ =(-1).\sum_{j=1}^{n}(-1)^{l+j}.a_{l,j}.|a(m\ |\ j)|\\ \color{red}{|E_{l,l+1}.A|=-|A|=|E_{l,l+1}|.|A|} $$

3.) $A=[a_{i,j}]$ and $B=E_{l,m}(k).A=[b_{i,j}]$ where $b_{m,j}=a_{m,j}+k.a_{l,j}$ $$ |E_{l,m}(k).A|=|B|=\sum_{j=1}^{n} (-1)^{m+j}.b_{m,j}.|b(m\ |\ j)|\\ =\sum_{j=1}^{n} (-1)^{m+j}.(a_{m,j}+k.a_{l,j}).|a(m\ |\ j)|\\ =\sum_{j=1}^{n} (-1)^{m+j}.a_{m,j}.|a(m\ |\ j)|+k.\sum_{j=1}^{n} (-1)^{m+j}.a_{l,j}.|a(m\ |\ j)|=|A|+0\\ \color{red}{|E_{l,m}(k).A|=|A|=|E_{l,m}(k).A|.|A|} $$

$\color{blue}{\text{Theorem 2}}:$ For the three possible elementary matrix we have the determinants, $$ \boxed{|E_{i,j}|=-1\\ |E_i{(k)}|=k\\ |E_{i,j}(k)|=1} $$ Proof :

Using $\color{blue}{\text{Theorem 1}}$,

$$ |E_{i,j}|=|E_{i,j}.I|=-|I|=-1\\ |E_i(k)|=|E_i(k).I|=k|I|=k.1=k\\ |E_{i,j}(k)|=|E_{i,j}(k).I|=|I|=1 $$

$\color{blue}{\text{Theorem 3}}:$ If $A$ is a nonsingular matrix, then there exists elementary matrices $E_1,E_2,E_3....E_n$ so that,

$$\boxed{ A=E_n....E_3.E_2.E_1.I=E_n....E_3.E_2.E_1} $$

$\color{blue}{\text{Theorem 4}}:$

A matrix $A$ is called singular if the equation $Ax=0$ has a nonzero solution.

Proof :

If $A$ is non-singular, ie. $|A|\neq0$, $$ Ax=b\implies A^{-1}\big(Ax\big)=A^{-1}b \quad (A \text{ is non-singular}\implies A\text{ is invertible}) \\Ix=A^{-1}b\implies x=A^{-1}b $$ Thus, $$ Ax=0\implies x=0 $$ If $A$ is non-singular $Ax=0$ can have only the trivial solution $x=0$. Thus, $$\boxed{ Ax=0 \text{ has non-trivial solution}\iff |A|=0 \text{ }\big(\text{i.e, }A\text{ is singular}\big)} $$

Case 1 : If $B$ is singular,

From $\color{blue}{\text{Theorem 4}}$,

there is a vector $x\neq 0$ such that $Bx=0$. Hence, $$ ABx=A(Bx)=A.0=0 $$ The equation $ABx=0$ also has a nontrivial solution $x\neq0$. Hence, $AB$ is singular ie. $|AB|=0$ $$ |AB|=0=|A||B| $$

Case 2 : If $B$ is non-singular,

$$ |AB|=|A.E_n....E_3.E_2.E_1|\qquad\color{blue}{\big(\text{Theorem 3}\big)} \\=|A|.|E_n|....|E_3|.|E_2|.|E_1|\qquad\color{blue}{\big(\text{Theorem 1}\big)} \\=|A|.|E_n....E_3.E_2.E_1|\qquad\color{blue}{\big(\text{Theorem 1}\big)} \\=|A|.|B|\qquad\color{blue}{\big(\text{Theorem 3}\big)}\qquad\color{white}{\big(\text{There.}\big)} $$

  • $\begingroup$ How do you show that $A$ singular implies $|A|=0$? $\endgroup$ – David C. Ullrich Feb 14 '18 at 15:12
  • $\begingroup$ @DavidC.Ullrich thanx for mentioning it. I have edited the answer. pls check. $\endgroup$ – ss1729 Feb 14 '18 at 17:25
  • $\begingroup$ Your answer is a much larger version than mine, but it essentially says the same thing ;) $\endgroup$ – Jose Lopez Garcia Feb 14 '18 at 17:50
  • $\begingroup$ I don't see how that answers my question at all. "Theorem 4" is more a definition than a theorem, and then in the proof you seem to be just assuming what I asked about. $\endgroup$ – David C. Ullrich Feb 14 '18 at 19:06
  • $\begingroup$ @DavidC.Ullrich srry i dont understand ur point. $A$ is singular means $|A|=0$ right ?. Thats wht we really mean by saying a matrix is singular. What u mean by "show that $A$ singular implies $|A|=0$" ? $\endgroup$ – ss1729 Feb 14 '18 at 19:35

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